Problem: $g(x) = -x^{2}-3x$ $h(t) = -5t^{3}-6t^{2}-7t-7-g(t)$ $ g(h(0)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(0)$ . Then we'll know what to plug into the outer function. $h(0) = -5(0^{3})-6(0^{2})+(-7)(0)-7-g(0)$ To solve for the value of $h$ , we need to solve for the value of $g(0)$ $g(0) = -0^{2}+(-3)(0)$ $g(0) = 0$ That means $h(0) = -5(0^{3})-6(0^{2})+(-7)(0)-7-0$ $h(0) = -7$ Now we know that $h(0) = -7$ . Let's solve for $g(h(0))$ , which is $g(-7)$ $g(-7) = -(-7)^{2}+(-3)(-7)$ $g(-7) = -28$